Can Evans's proof for the theorem regarding global approximation of Sobolev functions be significantly simplified? 

  Here $U$ is an open subset of $\mathbb{R}^n$.

Above is a theorem regarding approximation of Sobolev functions in Evans's Partial Differential Equations. When I tried to the recover the proof on my own, I found that the proof might be much shorter than the one in the book. But it looks too simple to be true and I'm wondering if there is a big gap there.

Here is my argument. Suppose $u\in W^{k,p}(U)$. Then $u\in L^p(U)$ by definition and thus $u\in L^1(U)$ since $U$ is bounded. Now according to the answer and comments to the following questions:  


*

*convolutions and mollification of functions in $L^1_{\text{loc}}(\Omega)$

*Properties of mollification for integrable functions
one can define the mollification $u^\epsilon=\eta_\epsilon*u$ on $U$ such that $u_\epsilon\in C^\infty(U)$. Moreover, since 
$$
D^\alpha u^\epsilon=\eta_\epsilon*D^\alpha u \quad\textrm{in  } U, 
$$
one has 
$$
\|u^\epsilon-u\|_{W^{k,p}(U)}^p=\sum \|D^\alpha u^\epsilon-D^\alpha u\|_{L^p(U)}^p\to 0.
$$

Could anyone identify if there is any serious mistake in the above argument?


[Added:]A possible naive analogy I make in the above argument is as the following. First of all, I have the following facts   


*

*$D^\alpha u^\epsilon=\eta_\epsilon*D^\alpha$ in $\color{blue}{U_\epsilon}$, where the definition can be seen in the linked question. 

*Also, $D^\alpha u^\epsilon\to D^\alpha u$ in $L^p(V)$ for any $V\Subset U$. 


Now that I can do mollification on the entire domain $U$ instead of just $U_\epsilon$, I just guess one might have


*

*$D^\alpha u^\epsilon=\eta_\epsilon*D^\alpha u$ in $\color{blue}{U}$.

*and $D^\alpha u^\epsilon\to D^\alpha u$ in $L^p(U)$. 



For the original "long" proof of THEOREM 2 by Evans, see this question. 
 A: NOTE. The numbering of equations is not progressive because of successive edits. I have left the original numbering so that comments stay understandable.

If you perform the mollification in the brutal way 
$$
u^\epsilon(x)= \int_{\mathbb R^n} \eta_\epsilon(x-y)u(y)\mathbf 1_{y\in U}\, dy,\quad x\in U $$ 
that is, setting $u$ to be equal to $0$ outside $U$, then the last two properties mentioned in your question need not hold in the whole domain $U$.
For a concrete example consider the space $W^{1,2}(-1, 1)$ and the constant function $u(x)=\mathbf 1_{|x|\le 1}.$ You have that 
$$\frac{du}{dx} \equiv 0\quad \text{in } U.$$ 
When you perform the brutal mollification you get 
$$\frac{d}{dx} \left( u\ast \eta_\epsilon\right)(x)= \eta_\epsilon(x+1) - \eta_\epsilon(x-1)\tag{2}$$
This already shows that the desired property 
$$\tag{!!!}\frac{d}{dx}\left( u\ast \eta_\epsilon\right) = \frac{du}{dx}\ast \eta_\epsilon\quad \text{in }U$$ 
needs not hold in $U$. Moreover, you can easily check that the $L^2(-1, 1)$ norm of (2) blows up to infinity as $\epsilon\downarrow 0$. This shows that the desired property 
$$\tag{!!!}\left
\| \frac{d}{dx}u\ast\eta_\epsilon - \frac{du}{dx}\right\|_{L^2(U)}\to 0$$ needs not hold either. 
Note that if you exclude a small neighborhood of $\{-1, +1\}$, that is, if you consider $U_\delta= (-1+\delta, 1-\delta)$ instead of $U$, then both properties $(!!!)$ do hold (the first holds provided $\epsilon < \delta$). That's why Evans worries about taking mollifications an epsilon away from the boundary of $U$.
Final remark. The distributional point of view provides some more insight. Note that, if you extend $u$ to be zero outside of $U$and consider the result as a distribution on $\mathbb R$, then its derivative is equal to 
$$\frac{ du}{dx} = \delta(x+1)-\delta(x-1).\tag{1}$$
The brutal extension produced two Dirac deltas in the first derivative. Those are responsible for the failure of properties $(!!!)$. 
